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Marrrta [24]
3 years ago
9

Pls help me!!!!!!!!!!!!!!!

Mathematics
1 answer:
Volgvan3 years ago
3 0
The answer is A because you have to subtract the pink ribbon by the red ribbon.
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Mrs. Lopez bought 6 rosebushes at Lowes for $22.08. What is the unit cost of one rosebush?
DedPeter [7]
It will be $3.68 for each rose bush because if she bought 6 of them for $22.08 then 22.08/6=3.68
8 0
3 years ago
What is the range of the function graphed below? PLEASE ANSWER ASAP
jenyasd209 [6]
<h3>Answer: Choice C) -\infty < y < 2</h3>

This is a more complicated way to write y < 2

The range is the set of all possible y outputs of a function. So we use the graph to see what y values are possible. The graph shows that y can be anything smaller than y = 2. We can't actually reach y = 2 itself due to a horizontal asymptote here.

In interval notation, the answer would be (-\infty, 2) with the curved parenthesis to indicate "do not include y = 2 as part of the range".

6 0
3 years ago
Determine the circumference of<br>a circle with diameter 28 cm. Show your work​
gulaghasi [49]

Answer:

Step-by-step explanation:

Diameter = 28 cm

therefore, radius = 28 / 2 = 14 cm

Circumference = 2 π r

= 2 × 22 / 7 × 14

= <u><em>88 cm</em></u>

Hope this helps

plz mark as brainliest!!!!!

8 0
3 years ago
Read 2 more answers
What is the value of b^2 0 4ac for the following equation?<br> 2x^2 0 2x 0 1 = 0<br> 04<br> 0<br> 12
tigry1 [53]
<span>2x^2 - 2x - 1 = 0 a = 2, b = -2, c = -1 b^2 - 4ac = (-2)^2 - 4(2)(-1) = 4 + 8 = 12
so the answer is 12
</span>
3 0
4 years ago
What is lim x→-3 sqrt x^2-8
vitfil [10]

If the  -8 is under the square root, then...

\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\

OR

If the -8 is not under the square root, then...

\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5

Either way, we replace x with -3 and simplify.

For more information, refer to the direct substitution rule for limits.

4 0
2 years ago
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